Vector A → has magnitude 8.78 m at 37.0 ∘ from the + x axis. Vector B → has magnitude 8.26 m at 135.0 ∘ from the + x axis. Vecto
1 answer:
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Answer:
(a) 0.42 m
(b) 20.16 N/m
(c) - 0.42 m
(d) - 0.21 m
(e) 17.3 s
Solution:
As per the question:
Mass, m = 0.56 kg
x(t) = (0.42 m)cos[cos(6 rad/s)t]
Now,
The general eqn is:
where
A = Amplitude
= angular frequency
t = time
Now, on comparing the given eqn with the general eqn:
(a) The amplitude of oscillation:
A = 0.42 m
(b) Spring constant k is given by:
Thus
(c) Position after one half period:
(d) After one third of the period:
(e) Time taken to get at x = - 0.10 m:
t = 17.3 s
Explanation:
The law of repulsion is given by Coulomb. The mathematical form of Coulomb law is given by :
...............(1)
Where
F is the force
k is the electrostatic constant
are electric charges
r is the distance between charges
The Newton's law of universal gravitation is given by :
..............(2)
G is the universal gravitational constant
From equation (1) and (2) it is clear that both law obeys inverse square law and both are of same type. So, the law of repulsion by Coulomb agrees with the Newton's law of universal gravitation.